Electromagnetic force on a particle in two different frames of reference

1. May 21, 2010

McLaren Rulez

Consider an infinitely long straight conductor carrying a current. Let's assume that the free charges in the conductor are positive and are moving at a drift velocity v. Now, consider a particle of charge +q also moving with v in the same direction as the current at a distance r from the conductor.

It faces two separate forces: One due to electrostatic repulsion from the charges in the wire and another because the wire has a magnetic field and a moving particle in a magnetic field experiences a force. The magnetic force points radially towards the wire and electrostatic repulsion points away from the wire.

Now, think of this in a frame which is moving at v in the same direction as the current. In this frame, the conductor no longer has a current; it is now just a straight piece of conductor with charge. The particle is also at rest in this frame. The only force it faces is electrostatic repulsion. The magnitude of this repulsive force does not change. So the two frames predict different answers to what is going to happen to the particle.

Where am I going wrong?

2. May 21, 2010

Staff: Mentor

You are going wrong right here:
Due to the relativistic effect of length contraction the charge density is higher and therefore the e-field is higher in the frame where the magnetic field disappears. Here is an interesting and easily readable page describing the relationship between relativity and magnetism.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

3. May 21, 2010

McLaren Rulez

Thanks Dale! I never thought about that.

So essentially, the phenomenon of magnetism is just a consequence of relativity and not a separate thing, is it?

Last edited: May 21, 2010
4. May 21, 2010

Staff: Mentor

Yes, you can obtain magnetism from electrostatics + relativity. Of course, it is kind of cheating since relativity was, in some sense, a result of Maxwell's equations.

5. May 21, 2010

Cleonis

Then again, it's not cheating at all because Maxwell's equations are a result of scientific effort to come up with the best possible description of Nature.

On a deeper level:
The Lorentz transformations were recognized as relevant long before relativistic physics was formulated. The Lorentz transformations are not unique to relativistic physics, in the sense that they embody something that is part of the fabric of Maxwell's equations.

In a sense relativistic physics was anticipated by Maxwell's equations. I find that fascinating. Maxwell's equations were conceived in terms of newtonian dynamics, yet they anticipate relativistic physics.

6. May 21, 2010

Staff: Mentor

Yeah, that sounds reasonable. Whether you start with electrostatics and relativity and derive magnetism or you start with electromagnetism and derive relativity really is a little irrelevant. You wind up with the same description of nature either way.

7. May 21, 2010

Bob S

Even if there is a current in the infinitely long conductor, I don't think there is any net charge in the conductor, so there is no net Coulomb force (to first order).

Bob S

8. May 21, 2010

Staff: Mentor

In the frame where there is no E-field that is correct. That is not correct in other frames. See the link I posted earlier, it is a very enjoyable read.

9. May 22, 2010

Bob S

Hi DaleSpam-
We agree that there is no transverse E-field (Coulomb) force in the lab frame. If we transform from the lab frame to the moving frame using the Lorentz transformations given in

http://pdg.lbl.gov/2002/elecrelarpp.pdf

(see third from last equation in the SI column), we get a pure transverse E-field force, as also shown in your referenced link Eqns (1)-(3):

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

which is an interesting alternative derivation of the Lorentz transformation for γ ≈ 1.

Bob S